Finding speed and length of a line

I have a math problem, which I'm kind of lost in. I have the question:

Find the speed $ds/dt$ on the line $x=1+6t$, $y=2+3t$, $z=2t$. Integrate to find the length $s$ from $(1,2,0)$ to $(13,8,4)$. Check by using $12^{2} + 6^{2} + 4^{2}$.

I think I have found the speed: $\sqrt{6^{2}+3^{2}+2^{2}} = 7$. But im lost in integrating to finding the length between the two points. Can anyone help me?

David

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If a particle travels along curve $s(t) = (1 + 6t, 2 + 3t, 2t)$ then its speed is $ds/ dt = (6, 3, 2)$.

The length of curve $s$ between two points $a = (1,2,0)$ and $b=(13, 8, 4)$ is computed as $\int_{t_a}^{t_b} \sqrt{(ds/dt)_x^2 + (ds/dt)_y^2 + (ds/dt)_z^2} dt$.

$t_a$ and $t_b$ are the points with the property $s(t_a) = a$ and $s(t_b) = b$. You need to determine them. Notice that $s(0) = a$ and $s(2) = b$.

With the values inserted, $$\int_{t_a}^{t_b} \sqrt{6^2 + 3^2 + 2^2} dt= \int_0^2 \sqrt{49} dt = \int_0^2 7 dt = 14$$

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Thanks man - helped a lot! – user1090614 Mar 3 '13 at 16:39
@user1090614 You are welcome! – goobie Mar 3 '13 at 16:41

Since this sounds like homework; hints:

• speed is simply the magnitude of the velocity.
• the magnitude of a vector $\vec{x}$ is $\sqrt{\vec{x} \cdot \vec{x}}$
• given a time-dependent (or constant) expression for the speed, if you integrate it over a period, you'll find the length of the distance travelled.
• i.e., you want an intermediate expression $s(t) = ...$, and in your case this expression will be very simple.
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Am I correct when stating s(t) = 7. And therefore integrating s(t) between 0 and 2 = 14? – user1090614 Mar 3 '13 at 16:31
sounds good to me... – Eamon Nerbonne Mar 3 '13 at 16:35
Super - I totally forgot integrating over a constant c = xc+d... Sometimes the answer is right infront of you, but your eyes wont let you see it. – user1090614 Mar 3 '13 at 16:37
Yeah, somehow particularly with math that's tricky. Wonder why... – Eamon Nerbonne Mar 3 '13 at 16:37